Question

Strontium nitrate, Sr(NO3)2, is used in fireworks to produce brilliant red colors, Suppose we need to prepare 366.6 ml. of 0.115 M SH(NO3)2 solution. How many grams of strontium
nitrate are required?
m(Sr(NO3)2) =

Answer 1

Answer:

$\boxed{\text{8.91 g}}$

Explanation:

1. Calculate the moles of Sr(NO₃)₂

$n = \text{366 mL} \times \dfrac{\text{0.115 mmol}}{\text{1 mL}}= \text{42.09 mmol}$

2. Calculate the mass of SrNO₃)₂

$m = \text{42.09 mmol} \times \dfrac{\text{211.63 mg}}{\text{1 mol}}= \text{8910 mg} = \text{8.91 g}\\\text{The mass of strontium nitrate required is }\boxed{\textbf{8.91 g}}$